1. ## Convergence

Does $\displaystyle{\prod_{\text{primes p}}\left(1+\frac{1}{p}\right)}$ converge, and if so to what value?

I'm not too sure how to go about this question, any help would be greatly appreciated!

2. The Swiss mathematician Leonhard Euler demonstrated two and half centuries ago that the series...

$\displaystyle \sum_{p} \frac{1}{p}$

... diverges, so that diverges also the product...

$\displaystyle \prod_{p} (1+\frac{1}{p})$

Kind regards

$\chi$ $\sigma$

3. However, if $s>1$ then,

$\displaystyle\prod_{p}\left(1+\dfrac{1}{p^s}\right )=\dfrac{\zeta(s)}{\zeta{(2s)}}$

So, can extend the formula,

$\displaystyle\prod_{p}\left(1+\dfrac{1}{p}\right)= \dfrac{\zeta(1)}{\zeta{(2)}}=\dfrac{+\infty}{\pi^2/6}=+\infty$

4. @chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?

@FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

thanks again!

5. Originally Posted by usagi_killer
@FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

We have for every $s>1$:

$\dfrac{\zeta (s)}{\zeta (2s)}=\dfrac{\prod_{p} (1-p^{-s})^{-1}}{\prod_{p} (1-p^{-2s})^{-1}}=\ldots$

6. Originally Posted by usagi_killer
@chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?
In...

Convergence of Infinite Products « The Everything Seminar

... there is the proof of the following theorem: an 'infinite product' of the form...

$\displaystyle \prod_{n=1}^{\infty} (1+a_{n})$

... converges if and only if the 'infinite sum'...

$\displaystyle \sum_{n=1}^{\infty} a_{n}$

... converges
...

Kind regards

$\chi$ $\sigma$

7. Thanks guys! Makes sense now!